3.550 \(\int \frac{A+B x^3}{(e x)^{7/2} \sqrt{a+b x^3}} \, dx\)

Optimal. Leaf size=246 \[ -\frac{\sqrt{e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} (2 A b-5 a B) \text{EllipticF}\left (\cos ^{-1}\left (\frac{\sqrt [3]{a}+\left (1-\sqrt{3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{5 \sqrt [4]{3} a^{4/3} e^4 \sqrt{\frac{\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{2 A \sqrt{a+b x^3}}{5 a e (e x)^{5/2}} \]

[Out]

(-2*A*Sqrt[a + b*x^3])/(5*a*e*(e*x)^(5/2)) - ((2*A*b - 5*a*B)*Sqrt[e*x]*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) -
a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*EllipticF[ArcCos[(a^(1/3) + (1 - Sqrt[
3])*b^(1/3)*x)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)], (2 + Sqrt[3])/4])/(5*3^(1/4)*a^(4/3)*e^4*Sqrt[(b^(1/3)*x*
(a^(1/3) + b^(1/3)*x))/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*Sqrt[a + b*x^3])

________________________________________________________________________________________

Rubi [A]  time = 0.183095, antiderivative size = 246, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {453, 329, 225} \[ -\frac{\sqrt{e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} (2 A b-5 a B) F\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt{3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{5 \sqrt [4]{3} a^{4/3} e^4 \sqrt{\frac{\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{2 A \sqrt{a+b x^3}}{5 a e (e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/((e*x)^(7/2)*Sqrt[a + b*x^3]),x]

[Out]

(-2*A*Sqrt[a + b*x^3])/(5*a*e*(e*x)^(5/2)) - ((2*A*b - 5*a*B)*Sqrt[e*x]*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) -
a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*EllipticF[ArcCos[(a^(1/3) + (1 - Sqrt[
3])*b^(1/3)*x)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)], (2 + Sqrt[3])/4])/(5*3^(1/4)*a^(4/3)*e^4*Sqrt[(b^(1/3)*x*
(a^(1/3) + b^(1/3)*x))/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*Sqrt[a + b*x^3])

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
 && FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(x*(s
+ r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[(r*x^2*(s + r*x^2))/(s + (1
+ Sqrt[3])*r*x^2)^2]), x]] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{A+B x^3}{(e x)^{7/2} \sqrt{a+b x^3}} \, dx &=-\frac{2 A \sqrt{a+b x^3}}{5 a e (e x)^{5/2}}-\frac{(2 A b-5 a B) \int \frac{1}{\sqrt{e x} \sqrt{a+b x^3}} \, dx}{5 a e^3}\\ &=-\frac{2 A \sqrt{a+b x^3}}{5 a e (e x)^{5/2}}-\frac{(2 (2 A b-5 a B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^6}{e^3}}} \, dx,x,\sqrt{e x}\right )}{5 a e^4}\\ &=-\frac{2 A \sqrt{a+b x^3}}{5 a e (e x)^{5/2}}-\frac{(2 A b-5 a B) \sqrt{e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} F\left (\cos ^{-1}\left (\frac{\sqrt [3]{a}+\left (1-\sqrt{3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{5 \sqrt [4]{3} a^{4/3} e^4 \sqrt{\frac{\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}\\ \end{align*}

Mathematica [C]  time = 0.0460112, size = 82, normalized size = 0.33 \[ -\frac{2 x \left (x^3 \sqrt{\frac{b x^3}{a}+1} (2 A b-5 a B) \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};-\frac{b x^3}{a}\right )+A \left (a+b x^3\right )\right )}{5 a (e x)^{7/2} \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/((e*x)^(7/2)*Sqrt[a + b*x^3]),x]

[Out]

(-2*x*(A*(a + b*x^3) + (2*A*b - 5*a*B)*x^3*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/6, 1/2, 7/6, -((b*x^3)/a)])
)/(5*a*(e*x)^(7/2)*Sqrt[a + b*x^3])

________________________________________________________________________________________

Maple [C]  time = 0.043, size = 3303, normalized size = 13.4 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/(e*x)^(7/2)/(b*x^3+a)^(1/2),x)

[Out]

2/5*(b*x^3+a)^(1/2)*(-4*A*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)
^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b
^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*
b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*x^5*b^3*e+10*B*(-(I*3^(1/
2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1
/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*
b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-
1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*x^5*a*b^2*e+10*B*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a
*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2
)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*
3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1
/2)-3))^(1/2))*(-a*b^2)^(2/3)*x^3*a*e+20*I*x^4*B*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/
2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a
*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-
1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*(-
a*b^2)^(1/3)*3^(1/2)*a*b*e+4*I*x^3*A*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/
2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-
2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))
/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*(-a*b^2)^(2/3)
*3^(1/2)*b*e-8*I*x^4*A*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1
/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)
^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2
)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*(-a*b^2)^(1/3)*3^(1/2)*b^2*e
+3*A*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3
)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*(-a*b^2)^(1/3)*((b*x^3+a)*e*x)^(1/2)*b-10*I*x^3*B*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(
1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^
2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*
EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(
1/2)+1)/(I*3^(1/2)-3))^(1/2))*(-a*b^2)^(2/3)*3^(1/2)*a*e+4*I*x^5*A*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-
a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/
2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I
*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(
1/2)-3))^(1/2))*3^(1/2)*b^3*e-10*I*x^5*B*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3
^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1
/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1
/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*3^(1/2)*a*
b^2*e+8*A*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a
*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I
*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/
2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*(-a*b^2)^(1/3)*x^4*b^2*e-20*B*(-(I*3^(1/2
)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/
2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b
^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1
+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*(-a*b^2)^(1/3)*x^4*a*b*e-4*A*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2)
)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(
1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*Elli
pticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)
+1)/(I*3^(1/2)-3))^(1/2))*(-a*b^2)^(2/3)*x^3*b*e-I*A*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3
)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*(-a*b^2)^(1/3)*3^(1/2)*((b*x^3+
a)*e*x)^(1/2)*b)/x^2/(-a*b^2)^(1/3)/b/a/e^3/(e*x)^(1/2)/((b*x^3+a)*e*x)^(1/2)/(I*3^(1/2)-3)/(1/b^2*e*x*(-b*x+(
-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))
)^(1/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{3} + A}{\sqrt{b x^{3} + a} \left (e x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/(e*x)^(7/2)/(b*x^3+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^3 + A)/(sqrt(b*x^3 + a)*(e*x)^(7/2)), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x^{3} + A\right )} \sqrt{b x^{3} + a} \sqrt{e x}}{b e^{4} x^{7} + a e^{4} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/(e*x)^(7/2)/(b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

integral((B*x^3 + A)*sqrt(b*x^3 + a)*sqrt(e*x)/(b*e^4*x^7 + a*e^4*x^4), x)

________________________________________________________________________________________

Sympy [C]  time = 176.107, size = 97, normalized size = 0.39 \begin{align*} \frac{A \Gamma \left (- \frac{5}{6}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{6}, \frac{1}{2} \\ \frac{1}{6} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt{a} e^{\frac{7}{2}} x^{\frac{5}{2}} \Gamma \left (\frac{1}{6}\right )} + \frac{B \sqrt{x} \Gamma \left (\frac{1}{6}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{6}, \frac{1}{2} \\ \frac{7}{6} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt{a} e^{\frac{7}{2}} \Gamma \left (\frac{7}{6}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)/(e*x)**(7/2)/(b*x**3+a)**(1/2),x)

[Out]

A*gamma(-5/6)*hyper((-5/6, 1/2), (1/6,), b*x**3*exp_polar(I*pi)/a)/(3*sqrt(a)*e**(7/2)*x**(5/2)*gamma(1/6)) +
B*sqrt(x)*gamma(1/6)*hyper((1/6, 1/2), (7/6,), b*x**3*exp_polar(I*pi)/a)/(3*sqrt(a)*e**(7/2)*gamma(7/6))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{3} + A}{\sqrt{b x^{3} + a} \left (e x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/(e*x)^(7/2)/(b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

integrate((B*x^3 + A)/(sqrt(b*x^3 + a)*(e*x)^(7/2)), x)